If a projectile is fired upward from the surface of the moon with an initial velocity of 40 m/s, its height, in metres, after t seconds is given by s=40t-.83t^2
a) Estimate the instantaneous velocity when t = a, (without using calculus)
b) Using your answer in part a, determine the maximum height.
Appreciate any help.
I don't know what the variable a and h are referring to ,
There were actually two questions before question a and b, and they were to find the average velocity between 1 and 3 seconds which is 38.705 m/s and to also find the instantaneous velocity at 2 seconds, which is 38.4763 m/s. Am I suppose to do something with the instantaneous velocity when t = 2?
In the earlier parts, they gave you a = 1 and a + h = 3 (so h = 2). Here you have a and a + h, so do the exact same steps as you did before.
The only difference will be that you won't be able to simplify as much, because you're working with a variable instead of which numbers.
The line in red is wrong. You have the function
.
(I wanted to use function notation so I added the parentheses and the t.) You plug in a + h for t:
Also, the line underneath the red one above isn't completely correct. You're not really saying that a equals (40a-.38a^2), but this is what the expression gives you when you plug in a. In function notation,
.
Now use the formula for the slope. The two points are (a, S(a)) and (a + h, S(a + h)):
Now take the limit as h -> 0:
For part b, set -1.66a + 40 equal to 0 and solve for a.
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